Question

solve
x^2+x+3=0 (mod 101)

Answer 1

Use the quadratic formula man! You keep posting the same type of questions over and over. The formula is:

(-b +/- SQRT(b^2-4ac)) / 2a

Answer 2

This is a congruence equation. (number theory)

Answer 3

hmm what do you mean mod 101? :-)

Answer 4

modulo

Answer 5

hmm |101| ?

Answer 6

basicly it is just for what x is x^2+x+3 divisible by 101

Answer 7

(there will be 0 or infinite many solutions)

Answer 8

oh okay cool stuff

Answer 9

But i dont know how to solve this one... First idea was to complete the square but that leaves the integers and that sucks :-)

Answer 10

hmm x^2 + x + 3
--------------- = k right?
101


k for integers now

(x + 1/2)^2 + 3 - 1/4 = k * 101

(x + 1/2)^2 = k*101 - 11/4 hmm maybe like this

Answer 11

\[x + \frac{1}{2} = \sqrt{k*101 - 11/4}\]

Answer 12

now for x to be valid \[k*101 > 11/4\]

Answer 13

\[x=\pm\sqrt{101n-\frac{11}{4}}-\frac{1}{2},\]\[\forall n\in\mathbb{Z}.\]What about this?

Answer 14

\[k > \frac{11}{4 \times 101}\]

Answer 15

x can only be integer as well!

Answer 16

oh I didn't knew about that sorry it's my first time solving these

Answer 17

number theory is all about integers

Answer 18

and divisibility also only makes sense with the integers

Answer 19

hmm k > 11/101*4
hmm and this is a small value so k must be valid for nearly every positive integer now

Answer 20

what I did may not be right but worth a thought: multiply by 34 to get
34(x^2+x)+102=0 (mod 101)
which is the same as
34(x^2+x)+1=0 (mod 101)
Or we can multiply by 67 (67*3=201) which is -1 mod 101
67(x^2+x)-1=0 (mod101)

Answer 21

Ok I will ask my seminar leader on Friday :-)